Max Value of r1+r2 in Inscribed Circles of Semicircle

[ritwik06]

Homework Statement

There is a semicircle with radius 1. Two circles are inscribe in it with centres C1 and C2 and radius r1 and r2 respectively. Find the maximum possible value of r1+r2
Here is the picture, I have drawn.
http://img143.imageshack.us/img143/8392/circlesdn3.png

The Attempt at a Solution

I have constructed the tangents as shown in the figure. I have shown the right angles. And I also deduce that:
AF=DF=CF
From this I see that r1 is necessarily equal to r2. Isnt it?? But I am not sure if I have drawn DF right. It might not be a tangent to both circles in some cases??

Help me!

 

[Dick]

I think I'm getting this. Pick the center of one of your circles. Can you show it lies on an ellipse? It's center is equidistant between the semicircle and the x-axis (the diameter of your semicircle). Pick C=(x,y) and find the conditions that imposes. Now concentrate on the quadrilateral formed by the centers of the two circles and the vertical lines dropping to the x-axis. That can give a condition relating the two circle radii. This is not an easy problem, as near as I can tell. It takes some work to put it together. Can you start this out? I'm only going to give hints.

 

[chaoseverlasting]

If the semi circle lies on the x axis, and if you have two circles with centers c1 and c2 with radii r1 and r2 (r1>r2) and you join the line joining the centers of the circles and extend it so that it intersects the x axis, and it makes an angle \theta with the x axis, you can come up with the following relation,

(r_1+r_2)sin\theta=r_1-r_2

where do you go from here?

 

[Avodyne]

To warm up, you could consider two special cases.

First, that the circles are of equal size, so that the point where they touch is on the y axis.

Second, that one circle is as large as possible, with radius 1/2 and its center on the y axis.

One of these two cases is likely to provide the maximum value of the sum of the radii, but actually proving that this gives the max value is harder. Dick's approach seems promising to me.